[英]Passing array variable to View in Codeigniter
美好的一天! 我有这段代码,我不知道我的代码可能是什么问题,因为我无法echo the value of variable in view
。
这是我的控制器:
public function generate()
{
$name = $this->input->post('name');
$this->data['generated'] = $this->user_models->generate_name($name);
$this->load->view('templates/header');
$this->load->view('generated_user_views', $this->data);
$this->load->view('templates/footer');
}
该模型:
function generate_name($name)
{
$this->db->select('*');
$this->db->from('user');
$this->db->where('name', $name);
$query = $this->db->get();
return $query->result_array();
}
在视图中:
<?php
$view_name = $generated['name'];
$view_department = $generated['department'];
?>
<div class="form-group">
<label for="name" class="col-sm-2 control-label emp">Name</label>
<div class="col-sm-10">
<input type="text" class="typeahead form-control" id="name" placeholder="Name" name="name" autocomplete="off"><?php echo $view_name; ?></input>
</div>
</div>
<div class="form-group">
<label for="department" class="col-sm-2 control-label emp">Department</label>
<div class="col-sm-10">
<input type="text" class="typeahead form-control" id="department" placeholder="Department" name="department" autocomplete="off"><?php echo $view_department; ?></input>
</div>
</div>
我不知道为什么我总是有这样的错误:
遇到PHP错误严重性:注意消息:未定义的索引:名称文件名:views / generated_user_views.php行号:12
遇到PHP错误严重性:通知消息:未定义的索引:部门文件名:views / generated_user_views.php行号:13
在模型中,您将从表中选择所有数据,因此返回了多维数组。在视图中,您需要遍历结果, 请参阅docs
foreach ($generated as $row)
{
echo $row['name'];
echo $row['department'];
}
或者我想如果您需要表中的一条记录,那么可以使用$query->row_array();
在模型中并在视图中可以按以下方式访问它们
$view_name =$generated['name'];
$view_department = $generated['department'];
尝试这个:
$name = $this->input->post('name');
$department = $this->user_models->generate_name($name);
$this->data = array('generated' => array('name' => $name, 'department'=> $department);
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